Problem: Let $y=\dfrac{5-3x}{x^2+3x}$. $\dfrac{dy}{dx}=$
Solution: $\dfrac{5-3x}{x^2+3x}$ is a rational expression. To find the derivative of rational expressions, we use the quotient rule : $\begin{aligned} \dfrac{d}{dx}\left[\dfrac{u(x)}{v(x)}\right]&=\dfrac{\dfrac{d}{dx}[u(x)]v(x)-u(x)\dfrac{d}{dx}[v(x)]}{[v(x)]^2} \\\\ &=\dfrac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2} \end{aligned}$ Let's differentiate! = d y d x = d d x ( 5 − 3 x x 2 + 3 x ) = ( x 2 + 3 x ) d d x ( 5 − 3 x ) − ( 5 − 3 x ) d d x ( x 2 + 3 x ) ( x 2 + 3 x ) 2 The quotient rule = ( x 2 + 3 x ) ( − 3 ) − ( 5 − 3 x ) ( 2 x + 3 ) ( x 2 + 3 x ) 2 Differentiate ( 5 − 3 x ) & ( x 2 + 3 x ) = − 3 x 2 − 9 x − ( 10 x + 15 − 6 x 2 − 9 x ) ( x 2 + 3 x ) 2 Expand = − 3 x 2 − 9 x − 10 x − 15 + 6 x 2 + 9 x ( x 2 + 3 x ) 2 = 3 x 2 − 10 x − 15 ( x 2 + 3 x ) 2 \begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}\left(\dfrac{5-3x}{x^2+3x}\right) \\\\ &=\dfrac{(x^2+3x)\dfrac{d}{dx}(5-3x)-(5-3x)\dfrac{d}{dx}(x^2+3x)}{(x^2+3x)^2} \gray{\text{The quotient rule}} \\\\ &=\dfrac{(x^2+3x)(-3)-(5-3x)(2x+3)}{(x^2+3x)^2} ~~~~\gray{\text{Differentiate }(5-3x)\text{ & }(x^2+3x)} \\\\ &=\dfrac{-3x^2-9x-(10x+15-6x^2-9x)}{(x^2+3x)^2} \gray{\text{Expand}} \\\\ &=\dfrac{-3x^2-9x-10x-15+6x^2+9x}{(x^2+3x)^2} \\\\ &=\dfrac{3x^2-10x-15}{(x^2+3x)^2} \end{aligned} In conclusion, $\dfrac{dy}{dx}=\dfrac{3x^2-10x-15}{(x^2+3x)^2}$, or any other equivalent form.